0
$\begingroup$

Given $T: \mathbb{R}^2 \to \mathbb{R}^2 : T\begin{bmatrix} x \\ y \end{bmatrix} \to \begin{bmatrix} 2x+y \\ x-3y \end{bmatrix}$ with standard basis $\mathcal{B}$ and basis $$\mathcal{B'}=\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \end{bmatrix} \right\}$$ Find the matrix $[T]_\mathcal{B'}^\mathcal{B'}$ representing the transformation $T$ with respect to the basis $\mathcal{B'}$ by making use of a change of basis matrix. Then calculate $[T]_\mathcal{B'}^\mathcal{B'}$ directly and check that you get the same answer.

Method 1

$T \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \qquad \quad T \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ -3 \end{bmatrix}$

$\Rightarrow [T]_\mathcal{B}^\mathcal{B} = \left[\begin{array}{rr} 2 & 1 \\ 1 & -3 \end{array}\right]$

To find the transition matrix $[I]_\mathcal{B}^\mathcal{B'}$ from $\mathcal{B}$ to $\mathcal{B'}$ I row reduced the augmented coefficient matrix $$\left[\begin{array}{cc|cc} 1 & 1 & 1 & 0 \\ 1 & 2 & 0 & 1 \end{array}\right] \rightarrow \left[\begin{array}{cc|rr} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & 1 \end{array}\right]^*$$ So $[I]_\mathcal{B}^\mathcal{B'}= \left[\begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array}\right]$ and $\left([I]_\mathcal{B}^\mathcal{B'}\right)^{-1}= \left[\begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array}\right]$

Now, $[T]_\mathcal{B'}^\mathcal{B'}= \left([I]_\mathcal{B}^\mathcal{B'}\right)^{-1} [T]_\mathcal{B}^\mathcal{B} [I]_\mathcal{B}^\mathcal{B'} = \left[\begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{rr} 2 & 1 \\ 1 & -3 \end{array}\right] \left[\begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array}\right] = \left[\begin{array}{rr} 8 & -5 \\ 13 & 9 \end{array}\right]$

Method 2

$T \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix} \qquad \quad T \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ -5 \end{bmatrix}$

$\Rightarrow [T]_\mathcal{B'}^\mathcal{B'}= \left[\begin{array}{rr} 3 & 4 \\ -2 & -5 \end{array}\right]$

Why are my solutions different?

$\endgroup$
1
  • $\begingroup$ I'm not going to accept an answer to this question in the foreseeable future because I don't understand linear algebra well enough to understand the correct answers. $\endgroup$ – ahorn May 24 '16 at 11:21
0
$\begingroup$

$T(1,1)$ and $T(1,2)$ should be expressed in basis $\mathcal B'$, not $\mathcal B$.

Let $\{e_1, e_2\}$ be the canonical basis. Set $u_1=\begin{bmatrix}1\\1 \end{bmatrix}=e_1+e_2$, $\,u_2=\begin{bmatrix} 1\\2\end{bmatrix}=e_1+2e_2$. From these, you deduce: $$e_1=2u_1-u_2,\quad e_2=u_2-u_1.$$

You've proved $\,T(u_1)= 3e_1-2e_2$, $\,T(u_2)=4e_1-5e_2$, whence: $$T(u_1)=8u_1-5u_2,\quad T(u_2)=13u_1-9u_2$$ so that $$T_\mathcal B^\mathcal{B'}=\begin{bmatrix}8&13\\-5&-9 \end{bmatrix}.$$

Incidentally, we proved the change of basis matrix from $\mathcal B'$ to $\mathcal B$ is $\,\begin{bmatrix}2&-1\\-1&1 \end{bmatrix}$.

$\endgroup$
8
  • $\begingroup$ Did I calculate my transition matrix in Method 1 correctly? I think that it is wrong, and I should have gotten the transition matrix simply by putting the basis vectors of $\mathcal{B'}$ together to form $\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$ $\endgroup$ – ahorn Apr 27 '15 at 10:18
  • $\begingroup$ The transition matrix I got in Method 1, $[T]_\mathcal{B}^\mathcal{B'}$, is supposed to be used by multiplying it with a vector in $\mathcal{B}$ to get a vector in $\mathcal{B'}$. Thus, it is not the transition matrix but rather the inverse. $\endgroup$ – ahorn Apr 27 '15 at 10:37
  • $\begingroup$ What exactly did I do wrong in method 2? $\endgroup$ – ahorn Apr 27 '15 at 10:42
  • $\begingroup$ Nothing is wrong. There's one more step to do. $\endgroup$ – Bernard Apr 27 '15 at 11:05
  • $\begingroup$ What does $\begin{bmatrix} 3 & 4 \\ -2 & -5 \end{bmatrix}$ represent? $\endgroup$ – ahorn Apr 27 '15 at 11:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.